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Asymptotic Analysis question.

  1. Jan 10, 2009 #1
    I have this question (it's not a coursework question, I'm learning it by my own):
    Use an appropiate power series expansion to find an asymptotic approximation as [tex]\epsilon[/tex] approaches 0+, correct to O(epsilon^2), for the two small roots of the equation: [tex]\epsilon x^3+x^2+2x-3=0[/tex] Then by using a suitable rescaling, find the first three terms of an asymptotic expansion as epsilon approaches 0+ of the singular root.


    My way to go around this is to write down the next expansion series (because of the degree of the equation):
    [tex]x(\epsilon)=\frac{1}{\epsilon^3}x_{-3}+ \frac{1}{\epsilon^2}x_{-2}+ \frac{1}{\epsilon^1}x_{-1}+x_0+x1\epsilon+...[/tex], the singular root is when epsilon equals null, i.e to solve x^2+2x-3=0=(x+3)(x-1) the roots are: x=-3 and x=1, I guess I need to choose the second root, now the recursive equation is:
    [tex]x^2_{n+1}=-x_{n}/\epsilon-2/\epsilon+3/(x_{n}\epsilon)[/tex] Now, x0=1 and then use the iterative equation in order to find x_-3 and x_-2, am I correct or totally way off here?

    Thanks in advance.
     
  2. jcsd
  3. Jan 11, 2009 #2
  4. Jan 11, 2009 #3

    lurflurf

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    h*x^3+x^2+2x-3=0
    h=-1/x-2/x^2+3/x^3
    take the leading term
    h=-1/x
    x=-1/h
    x=-1/h+2+ax+bx^2+...
    find more terms
     
  5. Jan 11, 2009 #4
    I don't quite understand what your method is... but if I were to tackle this problem, i'll get rid of the messy ϵ in x^3, and write:
    [tex]g(x)=f\left(\frac{1}{x}\right)=\frac{1}{x^3}(-3x^3+2x^2+x+\epsilon)[/tex]
    (the motivation is that as ϵ goes to 0, the extra root goes to infinity. This is not good. So plugging in 1/x makes the root go to zero instead).


    so, instead of finding roots of f, I find roots of g (much easier)
    g=0 implies
    [tex]x(3x^2-2x-1)=\epsilon[/tex]
    i'll just expand around -1/3, 1/2, 0 and solve this eq order by order.

    at the end, just take the inverse and expand using geometric series.
     
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